How is per-decision importance sampling derived in Sutton & Barto's book?
As mentioned in the comments your assumption about independence is wrong. Here's why. To prove independence we need to show the following holds:
$$P(X=x, Y=y) = P(X=x)P(Y=y)$$
in the case of RL this becomes:
$$P(X=a, X=a') = P(X=a)P(Y=a')$$
The left hand side has the value:
$$P(X=a, Y=a') = b(A_t = a| S_t = s) p(s'|a,s) b(A_{t+1} = a'|, S_{t+1} = s')$$
while the right hand side has the value:
$$P(X=a)P(Y=a') = b(A_t = a| S_t = s)b(A_{t+1} = a'| S_{t+1} = s')$$
And hence not independent.
Now let use look at why the following expression holds:

Eq.5.14:   $\mathbb{E}[\rho_{t:T-1}R_{t+k}] = \mathbb{E}[\rho_{t:t+k-1}R_{t+k}]$

I will not derive the exact expressions, but I hope you can form the reasoning I provide. By the rules of probability we know that sum of joint probability is equal to 1 i.e.:
$$\sum_{X_1..X_n} P(X_1=a_1, X_2=a_2,...X_n = a_n) = 1$$
I have alredy showed above, the trajectory is not independent. So $R_{t+k}$ will depend on the trajectory $S_{t:t+k-1}$ where $S_{t:t+k-1}$ is a particular trajectory. At the end of this trajectory we get a reward $R_{t+k}$ and thus $R_{t+k}$ is exclusively a function of $S_{t:t+k-1}$ i.e. $R_{t+k} = f(S_{t:t+k-1})$. The trajectory after this $S_{t+k:T-1}$ irrelevant since it will always sum up of to 1. i.e once you have reached a particular state at time step $t+k-1$ you are now conditioning based on that $P(S_{t+k:T-1}|S_{t:t+k-1})$ and taking the expected value over all trajectories possible from thereon i.e. $\sum_{S_{t+k:T-1}} P(S_{t+k:T-1}|S_{t:t+k-1})$ which is 1 by probability rules. Thus, what you are really doing is:
$$P(S_{t:t+k-1})R_{t+k}(\sum_{S_{t+k:T-1}} P(S_{t+k:T-1}|S_{t:t+k-1}))$$
and hence the remaining trajectory has no contribution.
Another way of thinking this is you are taking weighted trajectories till time step $t+k-1$ weighted by rewards $R_{t+k}$ and hence you cannot sum up to 1. The rest of the trajectory after $t+k-1$ will sum up to 1.
I hope this qualitative description suffices. You can do the maths, but you must be careful with the notations and the assumptions you make.
Also all the equations are correct, I hope you can indirectly see it from my reasoning.